1 solutions
-
0
C :
#include<stdio.h> #include<string.h> const int maxn = 3000; int main() { int k; int i, j; int f[8][maxn]; int num[3] = {0}; for(k = 0; k < 8; k++) { scanf("%d", &num[k]); } for(k = 0; k < 8; k++) { memset(f[k], 0, sizeof(f[k])); f[k][0] = 1; for(i = 2; i <= num[k]; i++) { int c = 0; for(j = 0; j < maxn; j++) { int s = f[k][j] * i + c; f[k][j] = s % 10; c = s /10; } } } for(k = 0; k < 8; k++) { for(j = maxn - 1; j >= 0; j--) { if(f[k][j]) break; } for(i = j; i >= 0; i--) { printf("%d", f[k][i]); } printf("\n"); } return 0; }
- 1
Information
- ID
- 504
- Time
- 1000ms
- Memory
- 2MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By